If a, b, c are three distinct real numbers in G.P. and a + b + c = xb,


If abc are three distinct real numbers in G.P. and a + b + c = xb, then prove that either x < −1 or x > 3.


Let $r$ be the common ratio of the given G.P.

$\therefore b=a r$ and $c=a r^{2}$

Now, $a+b+c=b x$

$\Rightarrow a+a r+a r^{2}=a r x$

$\Rightarrow r^{2}+(1-x) r+1=0$

$r$ is always a real number.

$\therefore \mathrm{D} \geq 0$

$\Rightarrow(1-x)^{2}-4 \geq 0$

$\Rightarrow x^{2}-2 x-3 \geq 0$

$\Rightarrow(x-3)(x+1) \geq 0$

$\Rightarrow x>3$ or $x<-1$ and $x \neq 3$ or $-1 \quad[\because a, b$ and $c$ are distinct real numbers $]$

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